How do collinear points on a matrix affect its rank? Consider the matrix 
\begin{matrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{matrix}
what effect does $({x_1},{y_1})$,$({x_2},{y_2})$,$({x_3},{y_3})$ being collinear have on the rank of the above matrix ?
 A: $$(x_i\,,\,y_i)\;,\,\,i=1,2,3\;,\;\;\text{are collinear}\;\;\iff \frac{y_3-y_2}{x_3-x_2}=\frac{y_3-y_1}{x_3-x_1}=\frac{y_2-y_1}{x_2-x_1}\iff$$ 
$$\iff (x_2-x_1)(y_3-y_2)=(y_2-y_1)(x_3-x_2)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_1)(y_2-y_1)=(x_2-x_1)(y_3-y_1)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_2)(y_2-y_1)=(x_2-x_1)(y_3-y_2)$$
If one of the denominators vanishes (and thus all of them), then $\,x_1=x_2=x_3\,$ , and we have two columns linearly dependent and the determinant in then zero, otherwise and using the above equality:
$$\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{vmatrix}=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=$$
$$=(x_3-x_2)(y_2-y_1)+x_2\left[(y_3-y_1)-(y_3-y_2)\right]+x_3(y_1-y_2)=$$
$$=0$$
so again zero. Do it slowly...!
A: Elementary row/column operations yield
$$
\left(\matrix{x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1}\right)\sim \left(\matrix{x_1&y_1&1\\x_2-x_1&y_2-y_1&0\\x_3-x_1&y_3-y_1&0}\right)\sim\left(\matrix{0&0&1\\x_2-x_1&y_2-y_1&0\\x_3-x_1&y_3-y_1&0}\right).
$$
So the rank of your matrix is the same as the rank of the latter, which is $1$ plus the rank of the lower left $2\times 2$ block.
Now your three points are collinear, if and only if the two vectors
$$
\vec{P_1P_2}=(x_2-x_1,y_2-y_1)\qquad\mbox{and}\qquad \vec{P_1P_3}=(x_3-x_1,y_3-y_1)
$$
are collinear. And this is equivalent to
$$
\det \left(\matrix{x_2-x_1&y_2-y_1\\x_3-x_1&y_3-y_1}\right)=0.
$$
Finally, the latter is equivalent to the fact that the rank of this $2\times 2$ matrix is $0$ or $1$.
So the rank of your initial matrix is $1$ or $2$ when the three points are collinear. And it is $3$ when they are not collinear.
